JS array to PHP and update table in MYSQL with PDO

Question

I have a form, where user can insert up to 4 age ranges. User can add or remove input fields dynamically.

I am trying to send the array to my PHP file and then insert it to database (insert new row for each range pair, lets say 1-3 and 4-6 will be on separate rows with correct user_id) with PDO.

The callback is success in PHP file, but nothing is inserted, which means it cannot get the values from JS and I am pretty lost how to send the array with the rest of the form data.

In html I have a button where, user can add fields dynamically and the ID will increment by 1 each time.

HTML file

<div class="row default" id="childAgeRange1">
    <div class="col-md-6">
        <input type="text" class="form-control" id="userProfileAgeFrom" name="userProfileAgeFrom[]" placeholder="Alates..." />
    </div>
    <div class="col-md-6">
        <input type="text" class="form-control" id="userProfileAgeTo" name="userProfileAgeTo[]" placeholder="Kuni..." />
    </div>
</div>

My JS file:

    $(document).on("click", "#btnUpdateInformation", function (e) {


                var userProfileAgeFrom = $("input[name^='userProfileAgeFrom']").map(function (idx, ele) {
                    return $(ele).val();
                }).get();

                var userProfileAgeTo = $("input[name^='userProfileAgeTo']").map(function (idx, ele) {
                    return $(ele).val();
                }).get();

                var formData = {
                    'user_id'                     : $("#hiddenUserID").val(),
                    'userPassword'                : $('#userPassword').val(),
                    'userRetypePassword'          : $('#userRetypePassword').val(),
                    'userBirthCountry'            : $('#userBirthCountry').val(),
                    'userBirthCity'               : $('#userBirthCity').val(),
                    'userBirthAddress'            : $('#userBirthAddress').val(),
                    'UserZipCode'                 : $('#UserZipCode').val(),
                    'userFirstName'               : $('#userFirstName').val(),
                    'userLastName'                : $('#userLastName').val(),
                    'userSex'                     : $('#userSex').val(),
                    'userBirthDay'                : $('#userBirthDay').val(),
                    'userBirthMonth'              : $('#userBirthMonth').val(),
                    'userBirthYear'               : $('#userBirthYear').val(),
                    'userPhoneNr'                 : $('#userPhoneNr').val(),
                    'userPasswordConfirm'         : $('#userPasswordConfirm').val(),
                    'userQuote'                   : $('#userQuote').val(),
                    'userDescription'             : $('#userDescription').val(),
                    'userProfileAgeFrom'          : userProfileAgeFrom,
                    'userProfileAgeTo'            : userProfileAgeTo,
                    'userProfileWage'             : userFinalWage

                };
        console.log(formData);
        $.ajax({
            type: "POST",
            url: "PHP/updateUserProfile.php",
            data: formData,  
            success: function(data){
                console.log(data);
                if(data.status == 'success'){
                    console.log("success");
                }else if(data.status == 'error'){
                    console.log("error");
                }else if(data.status == 'no_results'){
                    console.log("no results");
                }else if(data.status == 'results'){
                    console.log("there are results");
                }else if(data.status == 'password_matches'){
                    console.log("password matches");
                }

            },
            error: function(jqXHR, textStatus, errorThrown, data){
                console.log(jqXHR, textStatus, errorThrown, data);
            }
        });
        $('#btnUpdateInformation').unbind('click');
        e.preventDefault();
    });

PHP file

        if(isset($_POST['userProfileAgeFrom']) && isset($_POST['userProfileAgeTo'])){
            $data = array( 
                $userProfileAgeFrom => $_POST['userProfileAgeFrom'], 
                $userProfileAgeTo => $_POST['userProfileAgeTo']
            );

            $user_profile_age_range = $user_home->runQuery("UPDATE nanny_age_range SET age_minimum=:userProfileAgeFrom, age_maximum=:userProfileAgeTo WHERE user_id=:user_id ");
            $user_profile_age_range->bindparam(':userProfileAgeFrom', $userProfileAgeFrom, PDO::PARAM_STR);
            $user_profile_age_range->bindparam(':userProfileAgeTo', $userProfileAgeTo, PDO::PARAM_STR);
            $user_profile_age_range->bindparam(':user_id', $user_id, PDO::PARAM_STR);
            $user_profile_age_range->execute(); 
            $response_array['status'] = 'success';
        }

And database Table:

EDIT Console.log(formData) shows

UserZipCode
:
"11111"
userBirthAddress
:
"address"
userBirthCity
:
"c"
userBirthCountry
:
"c"
userBirthDay
:
"31"
userBirthMonth
:
"08"
userBirthYear
:
"1992"
userDescription
:
"description"
:
"x"
userLastName
:
"x"
userPassword
:
""
userPasswordConfirm
:
"xxxx"
userPhoneNr
:
"555555555555"
userProfileAgeFrom
:
"["1"]"
userProfileAgeTo
:
"["3"]"
userProfileWage
:
"9.60"
userQuote
:
"c"
userRetypePassword
:
""
userSex
:
"m"
user_id
:
"6"

@Cornwell var dump returns nothing, if I put it inside the function — z0mbieKale
– z0mbieKale, Commented May 24, 2016 at 8:33
Where are $userProfileAgeFrom, $userProfileAgeTo and $user_id defined? Your query also written as UPDATE rather than INSERT INTO. I clearly don't understand the question. — Chay22
– Chay22, Commented May 24, 2016 at 8:35
@Chay22, it should be update --- edited. The $userProfileAgeFrom and To comes from $_POST. They are defined like this $user_id = $_POST['user_id']; $userProfileAgeTo= trim($_POST['userProfileAgeTo']); $userProfileAgeFrom= trim($_POST['userProfileAgeFrom']); — z0mbieKale
– z0mbieKale, Commented May 24, 2016 at 8:39

vaso123 · Accepted Answer · 2016-05-24 14:55:40Z

1

Based on OP comment, console.log(formData) shows this:

userProfileAgeFrom: "["1"]"
userProfileAgeTo: "["3"]"

It means, now you have a JSON array in your values. So you need to decode it, and get the first value of it:

$userProfileAgeFrom = json_decode($_POST['userProfileAgeFrom'],true)[0], 
$userProfileAgeTo = json_decode($_POST['userProfileAgeTo'],true)[0]

EDIT

Based on OP comment.

So if you have more value in your userProfileAgeFrom and others, then iterate throught them:

for ($i = 0; $i < count($_POST['userProfileAgeFrom']); $i++) {
    $user_profile_age_range = $user_home->runQuery("UPDATE nanny_age_range SET age_minimum=:userProfileAgeFrom, age_maximum=:userProfileAgeTo WHERE user_id=:user_id ");
    $user_profile_age_range->bindparam(':userProfileAgeFrom', $_POST['userProfileAgeFrom'][$i], PDO::PARAM_STR);
    $user_profile_age_range->bindparam(':userProfileAgeTo', $_POST['userProfileAgeTo'][$i], PDO::PARAM_STR);
    $user_profile_age_range->bindparam(':user_id', $user_id, PDO::PARAM_STR);
    $user_profile_age_range->execute();
}

WARNING

But this will bad, because all your rows will be the same because of WHERE user_id=:user_id

I think you should redesign your code.

edited May 24, 2016 at 14:55

answered May 24, 2016 at 12:29

vaso123

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6 Comments

z0mbieKale Over a year ago

but what if there are more? there an be up to 4 different ranges (sorry for the late reply).

z0mbieKale Over a year ago

Also tried, it and getting this: Warning: json_decode() expects parameter 1 to be string, array given

z0mbieKale Over a year ago

any suggestions how can I edit the correct rows, so they all wont be the same?

vaso123 Over a year ago

create a hidden field for all fields, like <input type="hidden" name="userProfileAgeToOld[]" value="<?php echo $oldValue[$keyOfOldValuesArray];?>" /> and say WHERE userId = ... AND userProfileAge = userProfileAgeToOld[$i] AND ... So update by all values.

z0mbieKale Over a year ago

so if I add hidden input and append the row id from database to the value. can I then say like update where user_id = user_id and age_range_id = id?

|

Dinoop · Accepted Answer · 2016-05-24 11:08:13Z

0

The result of both $userProfileAgeFrom and $userProfileAgeTo are seems to array so the insertion will be work in a for loop. And hope the count of both this array are same.

$userProfileAgeFrom = $_POST['userProfileAgeFrom'];
$userProfileAgeTo = $_POST['userProfileAgeTo'];

for ($i = 0; $i < count($userProfileAgeFrom); $i++) {
    $user_profile_age_range = $user_home->runQuery("UPDATE nanny_age_range SET age_minimum=:userProfileAgeFrom, age_maximum=:userProfileAgeTo WHERE user_id=:user_id ");
    $user_profile_age_range->bindparam(':userProfileAgeFrom', $userProfileAgeFrom[$i], PDO::PARAM_STR);
    $user_profile_age_range->bindparam(':userProfileAgeTo', $userProfileAgeTo[$i], PDO::PARAM_STR);
    $user_profile_age_range->bindparam(':user_id', $user_id, PDO::PARAM_STR);
    $user_profile_age_range->execute();
}

answered May 24, 2016 at 11:08

Dinoop

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2 Comments

z0mbieKale Over a year ago

this solution does update the values in DB, but it updates the rows to the same result- lets say I typed in 1-3 and 5-7, both the rows will be 5-7

Dinoop Over a year ago

This is because you are try to run an update query with user_id. Try to provide auto increment id for the table nanny_age_range and update based on that id.

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